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Re: problems with the definition of a function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36669] Re: problems with the definition of a function
  • From: mnewstein at juno.com (Maurice Newstein)
  • Date: Wed, 18 Sep 2002 02:09:58 -0400 (EDT)
  • References: <am3n85$6tk$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

bagarell at unipa.it (fabio bagarello) wrote in message news:<am3n85$6tk$1 at smc.vnet.net>...
> Hi there!!
> I have quite an easy and annoying problem with mathematica:
> I need to define a function f(x,y) which takes some values for
> x=0,2pi,4pi (indepently of y) and has a different expression for all
> the other values of y. This is easily done for one-dimensional
> functions but I am in serious troubles for my two-dimensional problem:
> any suggestion?
> Thanks a lot,
> Fabio

Let the known values of the function at x={0,2Pi,4Pi} be {f0,f2,f4}.
Your conditions can be met by
f(x,y)=x(x-2Pi)(x-4Pi)g(y)+h(x),
where {h(0),h(2Pi),h(4Pi)}={f0,f2,f4}.
   An  h(x) can be found by quadratic interpolation:
 h[x]= a + bx + cx^2;
Solve[{a==f0,a+2Pi b +(2Pi)^2+(2Pi)^2 c==f2, a+4Pi b+(4Pi)^2 c==f4]
{{b -> -(3*f0 - 4*f2 + f4)/(4*Pi), 
  c -> -(-f0 + 2*f2 - f4)/(8*Pi^2), a -> f0}}

Maurice


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